One of the main uses of the Stein kernel is to be able to construct bounds on the Wasserstein distance between distributions on \(\mathbb {R}\)  (cf. Theorem 3.1 in [1]}). If we wish to adapt this theorem onto \(\mathbb {S}^1\)  for a circular distribution — say a von-Mises distribution — we will have to compute the Stein kernel. So far we have only looked at examples with the uniform measure on \(\mathbb {S}^1\)  due to its simple Lebesgue density.
However, for the von-Mises distribution in particular, one will quickly find that obtaining a closed form solution of the kernel is impossible.
For simplicity, let \(X\sim VM(0,\kappa )\) ; then
\( \tau (x) = \mathcal {T}_p(\mathbb {E}[X]-\mathrm {Id})(x) = e^{-\kappa \cos (x)}\int _{-\pi }^x (\mu -y)e^{\kappa \cos (y)} dy + \dfrac{C}{p(x)}. \) 
